Solving $t=q+pt^2$ with $q = 1-p$.

Question

I've come across something in my book that frustrates me. I feel like I should be able to solve this quickly but for some reason I don't know how. I want to find the roots $t$ to the equation $t=q+pt^2$, where $q = 1-p$. I know that she answers should be $t_1 = 1$ and $t_2 = q/p$, assuming that $p \neq 0$.

I end up with $$t = \frac{1 \pm \sqrt{1-4pq}}{2p}, $$ but can't figure out how to go from there... What am I not seeing about how to solve this?

Answer 1

You want to find a root of $f(t)=q-t+pt^2$. For $q=1-p$ this is $$f(t)=1-p-t+pt^2=(1-t)+p(t^2-1)=(t-1)[p(t+1)-1].$$ Hence, the roots are $t=1$ and $t=(1-p)/p=q/p$ provided that $p\neq0$.

Answer 2

$$1-4pq=1-4p(1-p)=1-4p+4p^2=(1-2p)^2$$ etc.

Answer 3

Note: $$t=q+pt^2 \Rightarrow t=1-p+pt^2 \Rightarrow t-1=p(t^2-1) \Rightarrow t-1=p(t-1)(t+1) \Rightarrow $$ $$t-1=0 \ \ \text{or} \ \ p(t+1)=1 \Rightarrow$$ $$t_1=1 \ \ \text{or} \ \ t_2=\frac{1}{p}-1=\frac{1-p}{p}=\frac{q}{p}.$$